Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Convolution of Two Monomials

  1. May 25, 2010 #1

    Char. Limit

    User Avatar
    Gold Member

    So I wanted to find the convolution of two monomials (don't ask me why) in the most general way I knew how. Of course, I only knew how for positive integer powers, but maybe someone can tell me how to do it for positive real powers (Gamma function maybe?). Anyway, here's what I did, replacing tau with s to make it easier to write:

    [tex]t^a \star t^b = \int_0^t s^a \left(t-s\right)^b ds[/tex]

    [tex]= \int_0^t s^a \left(t^b + \left(\left(-1\right)^1 b\right) s t^{b-1} + \frac{\left(-1\right)^2 b \left(b-1\right)}{2} s^2 t^{b-2} + ... + \frac{\left(-1\right)^{b-2} b \left(b-1\right)}{2} s^{b-2} t^2 + \left(\left(-1\right)^{b-1} b\right) t s^{b-1} + \left(-1\right)^b s^b\right) ds[/tex]

    [tex]= \int_0^t t^b s^a + \left(\left(-1\right)^1 b\right) t^{b-1} s^{a+1} + \frac{\left(-1\right)^2 b \left(b-1\right)}{2} t^{b-2} s^{a+2} + ... + \frac{\left(-1\right)^{b-2} b \left(b-1\right)}{2} t^2 s^{a+b-2} + \left(\left(-1\right)^{b-1} b\right) t s^{a+b-1} + \left(-1\right)^b s^{a+b} ds[/tex]

    [tex]= \frac{t^b s^{a+1}}{a+1} + \frac{\left(-1\right)^1 b t^{b-1} s^{a+2}}{a+2} + \frac{\left(-1\right)^2 b \left(b-1\right) t^{b-2} s^{a+3}}{2\left(a+3\right)} + ... + \frac{\left(-1\right)^{b-2} b \left(b-1\right) t^2 s^{a+b-1}}{2\left(a+b-1\right)} + \frac{\left(-1\right)^{b-1} b t s^{a+b}}{a+b} + \frac{\left(-1\right)^b s^{a+b+1}}{a+b+1} |_0^t[/tex]

    [tex]=\frac{t^{a+b+1}}{a+1} + \frac{\left(-1\right)^1 b t^{a+b+1}}{a+2} + \frac{\left(-1\right)^2 b \left(b-1\right) t^{a+b+1}}{2\left(a+3\right)} + ... + \frac{\left(-1\right)^{b-2} b \left(b-1\right)} t^{a+b+1}}{2\left(a+b+1\right)} + \frac{\left(-1\right)^{b-1} b t^{a+b+1}}{a+b} + \frac{\left(-1\right)^b t^{a+b+1}}{a+b+1}[/tex]

    [tex]= t^{a+b+1} \left(\frac{\left(-1\right)^0 b!}{\left(a+1\right) 0! \left(b-0\right)!} + \frac{\left(-1\right)^1 b!}{\left(a+2\right) 1! \left(b-1\right)!} + \frac{\left(-1\right)^2 b!}{\left(a+3\right) 2! \left(b-2\right)!} + ... + \frac{\left(-1\right)^{b-2} b!}{\left(a+b-1\right) \left(b-2\right)! 2!} + \frac{\left(-1\right)^{b-1} b!}{\left(a+b\right) \left(b-1\right)! 1!} + \frac{\left(-1\right)^b b!}{\left(a+b+1\right) \left(b-0\right)! 0!} \right)[/tex]

    [tex]t^a \star t^b = t^{a+b+1} \sum_{n=0}^b \frac{\left(-1\right)^n b!}{\left(a+n+1\right) n! \left(b-n\right)!}[/tex]

    Does this seem right? And is it possible to generalize this to powers other than positive integers?
    Last edited: May 25, 2010
  2. jcsd
  3. May 27, 2010 #2

    Char. Limit

    User Avatar
    Gold Member

    Bumping, it's been two or three days.

    I do want to know one thing: the formula I have only holds for positive integer b. Assuming everything I did holds to be true, can I simply change the factorials to gamma functions (changing the b to b-1 of course) and continue from there?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook